Análisis cualitativo

Tadeusz Górecki Ionic Equilibria

Page 20

Acid-Base Equilibria

Brønsted-Lowry: an acid is a proton _______, a base is a ______________.++↔ HBaseAcid

________________ (H3PO4, H2O), ______ ( +4NH ) and _____ (H2PO4

-) can allbehave as acids.

Example:++ +↔ HNHNH 34

Substances which can behave both as acids and as bases: ____________, or______________ substances (e.g. H2O, SH-).

baseacidSHSH −+− +↔ 2

acidbaseSHSHH 2↔+ −+

Free protons _________ in any solvent, thus the above reactions are______________. In reality:

++ +↔+ OHNHOHNH 3324

Energy required to dissociate _____ to ____ and __: _____ kcal/mol


1

Equilibrium constant for _______________:−+ +↔+ AOHOHHA 32

][]][[

HAAHKa

−+

=

________________:−+ +↔+ OHBHOHB 2

][]][[

BOHBHKb

−+

=

Relationship between Ka and Kb:

wba KKK =⋅

b

wa K

KK =a

wb K

KK =

Lewis: an acid is an ___________________; a base is an _________________.

________________________________________________________________

Strength of acids and bases−+− +↔+ 2

4324 SOOHOHHSO

−+ +↔+ 33232 HCOOHOHCOH

−+ +↔+ CNOHOHHCN 32

1221 baseacidbaseacid +↔+

_______][

]][[

4

243 == −

−+

HSOSOOHKa

_______][

]][[

32

33 ==−+

COHHCOOHKa


Page 22

______][

]][[ 3 ==−+

HCNCNOHKa

_______ value of __ means that the acid is ________, thus:

______________________

____________ of water:−+ +↔ OHHOH2

Equilibrium constant using __________:

OH

OHH

aaa

K2

0 −+=

Activity of water is by thermodynamic convention proportional to the _____________ of water in the solution. In dilute solutions it is close to __.

Activity of water can be _____________________:

___][][ == −−

++

−+ γγ OHHaa OHH

"________________" constant:

wKOHH =−+ ]][[

−+−+

== γγγγ

/00 2w

OHw K

aKK


Page 23

At 50°C, pKw = ______, and the neutral point is pH = ____. At 25°C in 3 MNaClO4 pKw = _____, and the neutral point is pH = ____.


Page 24

____________ solvents:−+ +↔+ 2433 NHNHNHNH

At -60°C, the equilibrium constant is:32

24 10]][[ −−+ == NHNHK

Thus, the pH scale (defined as -log[NH4+]) in liquid ammonia ranges from _ to

__.

pH of a strong acid

Initially PH, or "_________________", defined as

PH = -log CH

Today's definition of pH:

)]log([log ++−=−== γHapapH HH

General approach

Example: HCl

Mass balance: ____][ =−ClIon product of water: ____]][[ ==−+

wKOHHCharge balance: ][][][ −−+ += ClOHH

Solution: HAw C

HKH += +

+

][][

This is a quadratic equation, which applies ________.

When ____________, [H+] = ___ ([OH-] is _______________)

At higher ionic strength, activity coefficient should be used.


Page 25

Strong base:

Example: NaOH

Mass balance: ___][ =+NaIon product of water: 1410]][[ −−+ == wKOHHCharge balance: _____________________

Solution:][

][ ++ =+

HKCH w

b

Basic solution, thus ______________, and in general

b

w

CKH =+ ][

___________________

____________________________________

Example: pH of 7102 −⋅ M solution of NaOH

0][][ 2 =−+ ++wb KHCH

24

][2

wbb KCCH

++−=+

LmolH /1014.4][ 8−+ ⋅= pH = _____

Simplified equation: pH = _____

____________________________________


Page 26

pH of strong acid/base as a function of concentration:

-10-9-8-7-6-5-4-3-2-10

0 2 4 6 8 10 12 14

pH

log

Clog C(acid) log C(base)

Mixture of a strong acid and a strong base

Example: HCl and NaOH

Mass balance: ___][ =−ClMass balance: ___][ =+NaIon product of water: 1410]][[ −−+ == wKOHHCharge balance: ][][][][ −−++ +=+ OHClNaH

][][ +

+ −=−HKHCC w

ba

When the acid and the base are ____________:

LmolKOHH w /10][][ 7−−+ ===


Page 27

Titration of Strong Acids and Bases

Volume of the system changes, thus _______ must be taken into mass balancesrather than _______________.

Example: titration of HCl with NaOH:

Mass balance: ________)]([ =+−ba VVCl

Mass balance: ________)]([ =++ba VVNa

Ion product of water: 1410]][[ −−+ == wKOHHCharge balance: ][][][][ −−++ +=+ OHClNaH

][][ +

+ ++

=+

+HK

VVVC

VVVCH w

ba

aa

ba

bb

At the equivalence point, __________ (1:1 stoichiometry) and ___________

Before the equivalence point, _______________:

ba

bbaa

VVVCVCH

+−

=+ ][

After the equivalence point, _______________:

ba

aabbw

VVVCVC

HK

+−

=+ ][

aabb

baw

VCVCVVKH

−+

=+ )(][


Page 28

In the vicinity of the equivalence point (Ca = 0.1 M, Va = 50 mL, Cb = 0.2 M):

5

6

7

8

9

24.999 24.9995 25 25.0005 25.001

Vb

pH

OH- neglectedFull equationH+ neglected

Plotting the titration curve

__________ titrated with _________:

−++−

= ++

++

]/[][]/[][

HKHCHKHCVV

wb

waab

Titration of ___________ with ___________:

+−−+

= ++

++

]/[][]/[][

HKHCHKHCVV

wa

wbba


Page 29

Example:

Titration curve

0

2

4

6

8

10

12

14

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50

Vb

pH

_______________ titration:

HCl titrated with NaOH

][][][][ −−++ +++=Λ ClOHNaH ClOHNaH λλλλ

Λ - _______ conductance ( ][ 11 −−Ω cmk )Xλ - __________ conductance.

At 25°C, ________ conductances 0λ are:

_____0 =+ λH_____0 =+ λNa_____0 =− λOH_____0 =− λCl

Procedure:

• ______ the values of pH;


Page 30

• ________ [H+] and [OH-];• calculate V from the ____________ equation;• calculate )/(][ 0VVVCNa b +=+ (from ___________);

• calculate )/(][ 00 VVVCCl a +=− (from ___________).

50 mL 0.1 M HCl titrated with 0.2 M NaOH:

0

5

10

15

20

25

30

35

0 10 20 30 40 50

Vb

Con

duct

ance

50 mL 0.0001 M HCl titrated with 0.0001 M NaOH:

0.006

0.0065

0.007

0.0075

0.008

0.0085

0.009

0.0095

0.01

0.0105

0.011

40 50 60

Vb

Con

duct

ance


Page 31

Titration error

''

VVV

errorTitration ep −=

Vep - V at ________V' - V at _______________Titration of 50 mL 0.1 M HCl with 0.2 M NaOH:

Enlarged section (end point detected with __________ at pH = 5):


Page 32

Titration error:

_________25

259963.24100 =−

⋅

________________50 mL 0.0001 M HCl titrated with 0.0001 M NaOH:

Titration error:

______50

5040100 =−

⋅

Gran plotsTitration of a ___________ with a __________:

][][ +

+ ++

=+

+HK

VVVC

VVVCH w

ba

aa

ba

bb

Before the equivalence point, ______ is negligibly small, thus:

bbaaba VCVCHVV −=+ + ])[(


Page 33

or bbaapH

ba VCVCVVf −=+= −10)(1

__ , ___ and __ are _________, thus a plot of f1 as a function of Vb should be a____________ with a slope of _____ intersecting the X axis at the _______________, _____________.Example: 50 mL 0.0001 M HCl titrated with 0.0002 M NaOH:

0

0.00005

0.0001

0.00015

0.0002

0.00025

0.0003

0.00035

0.0004

0.00045

0.0005

22 23 24 25 26 27

Vb [mL]

f1

In the vicinity of the equivalence point:

0

0.00001

0.00002

0.00003

0.00004

0.00005

0.00006

0.00007

0.00008

0.00009

0.0001

24.5 24.6 24.7 24.8 24.9 25 25.1 25.2 25.3

Vb

f1


Page 34

Weak monoprotic acids and bases

−+ +↔ AHHA

][]][[

HAAHKa

−+

=

−+ +↔+ OHBHOHB 2

][]][[

BOHBHK b

−+

=

wba KKK =⋅

wba pKpKpK =+


Page 35

NH4+: __________

NH3: __________Dependence of ___ on ___________:

00

0

][ ]][[

γγγ

γγγ −+−+

−+

== aa KHA

AHK

00 logloglog γγγ −++= −+aa pKpK

Using _______________ and setting bI=0logγ (activity coefficient for an______________________):

bIIbI

IpKa −

−

+⋅−= '

151.02757.4

where b' is the ________________ (usually 0.2).


Page 36

Best fit: _______________

Temperature dependence of apK 0 :


Page 37

Calculating the pH of weak acidKnown: ___________Unknown: ______________________

][]][[

HAAHKa

−+

=

]][[ −+= OHHK w

Mass balance: ][][ HAACHA += −

Charge balance: ][][][ −−+ += OHAH

+=

+=+=

+−

+−

−+−

a

a

aaHA K

KHA

KHA

KAHAC

][][][1][]][[][

a

aHA

KHKC

A+

=+

−

][][

From ______________:

a

HA

KHHCHA+

= +

+

][][][

From ____:

][][ +

− =HKOH w

Substituting ____ and _____ into __________________:

][][][ ++

+ ++

=HK

KHKCH w

a

aHA

Thus:

0)]([][][ 23 =⋅−+−+ +++wawaHAa KKKKCHKHH


Page 38

Simplifying assumption: _____ is negligibly small

a

aHA

KHKCH+

= ++

][][

0][][ 2 =−+ ++aHAa KCKHH

When ____________:

][][ +

+ =H

KCH aHA aHAKCH =+ 2][ aHAKCH =+ ][

Flood's diagramFrom ____ and _______________:

−

+= +

++

][][][

HKH

KKHC w

a

aHA

Flood's diagram

-8

-6

-4

-2

00 2 4 6 8

pH

log C

Strong acidpKa=4.75pKa=7.53pKa=10.72


Page 39

Degree of dissociation

a

aHA

KHKCA+

= +−

][][

a

HA

KHHCHA+

= +

+

][][][

Degree of _____________:

=+

== −

−−

][][][][AHA

ACA

HA

α

Degree of ____________:

=+

==− − ][][][][1AHA

HACHA

HA

α

Degree of dissociation and formation

0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10 12 14

pH

degree of dissociationdegree of formation


Page 40

Sillén's diagram (______________, ____________________)Acetic acid, 0.01 M, pKa = 4.75

-14

-12

-10

-8

-6

-4

-2

00 2 4 6 8 10 12 14

pH

log

C

[OH-] [H+]

[A-]

[HA]

1. [H+] is determined from the _______________:

pHH −=+ ]log[

2. [OH-] is determined from _________________:

ww pKpH

HKOH −== +

−

][log]log[

3. [A-] is determined from ____________ and ___:

a

aHA

KHKCA+

= +−

][][

4. [HA] is determined from ____________ and ___:

a

HA

KHHCHA+

= +

+

][][][


Page 41

pH of a given system can be determined from the _______________:

][][][ −−+ += OHAH

Acidic solution, thus _____ can be neglected

][][ −+ = AH

Solution for the proton condition can be easily found on equilibrium diagramsusing the _______________:

( )][][][log −−+ −− OHAH

Acetic acid, _____ M, pKa = _____

3.4

-14

-12

-10

-8

-6

-4

-2

00 2 4 6 8 10 12 14

pH

log

C

[OH-] [H+]

[A-]

[HA]

Pointer


Page 42

Acetic acid, ____ M, pKa = ____

6.8

-14

-12

-10

-8

-6

-4

-2

00 2 4 6 8 10 12 14

Pointer

[OH-][H+]

[A-][HA]

Plotting equilibrium diagrams

1. [H+] and [OH-]: _______ lines at ________ (slopes of __ and __,respectively)

2. [A-]:

a

aHA

KHKCA+

= +−

][][

for ____________________

][][ +

− ≅H

KCA aHA

pHpKCA aHA +−=− log]log[

1]log[=

−

dpHAd

for ___________________ HACA ≅− ][


Page 43

3. [HA]:

a

HA

KHHCHA+

= +

+

][][][

for ___________________ HACHA =][

for ___________________

a

HA

KHCHA ][][

+

≅

aHA pKpHCHA +−= log]log[

1]log[−=

dpHHAd

4. When __________:

2][][ HACHAA ==−

3.0log2loglog2

log]log[]log[ −=−===−HAHA

HA CCCHAA

What is the pH of 0.001 M NaAc?

____:

][][][ −+ =+ OHHAH

___________, thus _____:

][][ −= OHHA


Page 44

0 2 4 6 8 10 12 14 -14

-12

-10

-8

-6

-4

-2

0

pH

log

Con

cent

ratio

n [M

]


Page 45

What happens when the acid concentration is __________?

0.001 M HF, pKa = 3.17

Proton condition: ][][][ −−+ += OHFH

][][ −+ ≅ FH

-14

-12

-10

-8

-6

-4

-2

00 2 4 6 8 10 12 14

pH

log

C

[OH-][H+]

[F-]

[HF]


Page 46

-14

-12

-10

-8

-6

-4

-2

00 2 4 6 8 10 12 14

pH

log

CPointerpH=3.26

[OH-][H+]

[F-]

[HF]

Checking the results:

pH = ____

[OH-] = ______

[HF] = ______

=⋅

== −

−−−+

34.3

26.326.3

101010

][]][[

HFFHKa

Mass balance:

=+=+ −−− 34.326.3 1010][][ HFF

Algebraic solution:

0][][ 2 =−+ ++aHAa KCKHH

=+ ][H

pH = ____


Page 47

Low concentrations of very weak acids:

5 x 10-5 M HCN, pKa = 9.32

-14

-12

-10

-8

-6

-4

-2

00 2 4 6 8 10 12 14

pH

log

C[OH-][H+]

[HCN] [CN-]

[CN-]+[OH-]

Mixture of acids

Strong acids represented by ______________.

Strong bases represented by ______________.

Typically ___________ on EquiligrapHs.

Each __________ or _____ represented by the expressions:

a

aHA

KHKCA+

= +−

][][

a

BH

KHHCBH+

= +

++

][][][

pH found at the point where _______________.


Page 48

0.01 M HAc (Ka = 10-4.75) and 0.001 M HFo (Ka = 10-3.75)

-14

-12

-10

-8

-6

-4

-2

00 2 4 6 8 10 12 14

pH

log

C

[OH-] [H+]

[Ac-][HAc]

[HFo] [Fo-]

[Ac-]+[Fo-]+[OH-]

Proton condition: _________________________

Pointer function:

-14

-12

-10

-8

-6

-4

-2

00 2 4 6 8 10 12 14

pH

log

C

PointerpH=3.3

[OH-] [H+]

[Ac-][HAc]

[HFo] [Fo-]


Page 49

Assumptions:

______ negligibly small

__________

][][][][ +++

+ ++

++

=HK

KHKC

KHKCH w

f

ff

a

aa

f

ffaa

KHKC

HKCH

++= ++

+

][][][

Iteration:f

ffaa KH

KCHKCH

++= +

++

][][

][ 2

Circular reference:f

ffaa KH

KCHKCH

++= +

++

][][

][

Mixture of strong and weak acid:

0.001 M HCl and 0.01 M HAc

][][][][ −−−+ ++= OHClAcH

][][

][][ +

−+

+ +++

=HKCl

KHKCH w

a

aa

____________________

]][[][ 2 −++ += ClHKCH aa

Solution: pH = _____


Page 50

-14

-12

-10

-8

-6

-4

-2

00 2 4 6 8 10 12 14

pH

log

C

[OH-] [H+]

[Ac-]

[HAc]

[Cl-]

[Ac-]+[Cl-]+[OH-]

Salt of a weak acid and a weak base

Two independent ________________ linked by the condition that they have thesame ___________________.

][]][[ 1 HAKAH a=−+

][]][[ 2++ = BHKBH a

wKOHH =−+ ]][[

____ balances: ][][][][ BBHAHAC +=+= +−

______ balance: ][][][][ −−++ +=+ OHAHBH

_______________: ][][][][ BOHHHA +=+ −+


Page 51

If [H+] and [OH-] _________:

2

2

1 ][][][

a

a

a KHCK

KHHC

+=

+ ++

+

212][ aa KKH ⋅=+

21][ aa KKH ⋅=+

pH _______________ on C (provided the assumption above is fulfilled)!

Example: pH of 0.01 M NH4Ac (pKa1 = 4.75, pKa2 = 9.25)

=+= )(2/1 21 aa pKpKpH

The value of __ is coincidental.

Equilibrium diagram:

-14

-12

-10

-8

-6

-4

-2

00 2 4 6 8 10 12 14

pH

log

C

[OH-] [H+]

[Ac-]

[HAc][NH3]

[NH4+]


Page 52

Pointer function:

-14

-12

-10

-8

-6

-4

-2

00 2 4 6 8 10 12 14

pH

log

C

PointerpH=7

[OH-] [H+]

[Ac-]

[HAc][NH3]

[NH4+]

Full solution:

][][][

][][

2

2

1++

++

+

++

=++ H

KKH

CKHKH

HC w

a

a

a

_______ equation in [H+].

The equation is _____________, thus:

+

−

+

+

−=

+++

++

2

2

1 ][1

][1][

][][

a

a

a

w

KHK

KHH

HHK

C


Page 53

Example: Dimethylammonium acetate

pKa1 = 4.75 (___________), pKa2 = 10.76 (__________________)

7

7.1

7.2

7.3

7.4

7.5

7.6

7.7

7.8

0 2 4 6 8-log C

pH

pH = 7.755

General equation for the titration curve

____ balance:ba

aa

VVVCAHA+

=+ − ][][

____ balance:ba

bb

VVVCNa+

=+ ][

][]][[

HAHAKa

+−

=

______ balance: ][][][][ −−++ +=+ OHANaH


Page 54

a

a

ba

aa

KHK

VVVCA

+⋅

+= +

−

][][

][][][ ++

+ ++

⋅+

=+

+HK

KHK

VVVC

VVVCH w

a

a

ba

aa

ba

bb

or][

][ ++ +

+⋅=

++

HK

VVVC

VVVCH w

ba

aaHA

ba

bb α

−+

+−

+⋅

=

++

++

+

][][

][][

][

HKHC

HKH

KHKC

VVw

b

w

a

aa

ab

−+

+−⋅=

++

++

][][

][][

HKHC

HKHC

VVw

b

wHAa

ab

α

Titration of a ___________ with a __________:

−+

+−=

++

++

][][

][][

HKHC

HKHC

VVw

b

wa

ab

Simplifying assumptions: before the equivalence point ____________; after theequivalence point ________________


Page 55

Example: titration of 10 mL 0.1 M HAc with 0.1 M NaOH

0

2

4

6

8

10

12

14

0 2 4 6 8 10 12 14Vb

pH

________________

)()(

)()(

bb VpH

VpH

∆∆

≈∂∂

End point determined by the _________ of the ______________. If necessary,________________ can be obtained in the same manner.

________________:

0

2

4

6

8

10

12

14

0 2 4 6 8 10 12 14Vb

pH

d(pH)/dVb


Page 56

____ and ________________:

-8

-6

-4

-2

0

2

4

6

8

10

12

14

9.90 9.95 10.00 10.05 10.10

Vb

Another way to plot the titration curve: through _____________ Φ :

aa

bb

VCVC

=Φ

Substitution to the general equation:

−+

++

=Φ +++ ][][

)(][

HHK

VCVV

KHK w

aa

ba

a

a

Iteration – set ________________ and get the ________________


Page 57

0

2

4

6

8

10

12

14

0 1 2Fraction titrated

pH

_______________ solution:

b

a

aa

bb

COHH

COHH

VCVC

][][1

][][

−+

−+

−+

−−

==Φα

0

2

4

6

8

10

12

14


pH


Page 58

The plot of _____________________ enables easy comparisons of different_______________.

Examples:

• Different _________ of the acid

0

2

4

6

8

10

12

14


pH

• Different ___ values

0

2

4

6

8

10

12

14


pH

Ka = 1e-6Ka = 1e-5Ka = 1e-4

Ca = Cb = 0.01 M

0

2

4

6

8

10

12

14

0 5 10 15 20 25 30 35 40Vb

pH

Va = 10 mLVa = 20 mL

Va = 30 mL

0

2

4

6

8

10

12

14

0 5 10 15 20Vb

pH

Ka = 1e-6Ka = 1e-5Ka = 1e-4

Ca = Cb = 0.01 M


Page 59

• Different __________________________:

0

2

4

6

8

10

12

14

0 1 2

Fraction titrated

pH

Ca = 0.0001 M

Ca = 0.001 M

Ca = 0.01M

pKa = 4.75Cb = 0.01 M

• Different concentrations of the _____:

0

2

4

6

8

10

12

14

0 1 2

Fraction titrated

pH

Cb = 0.0001 M

Cb = 0.001 M

Cb = 0.01 M

pKa = 4.75Ca = 0.01 M

0

2

4

6

8

10

12

14

0 5 10 15Vb

pH

Ca = 0.0001 M Ca = 0.001 M

Ca = 0.01 M

pKa = 4.75Cb = 0.01 M

0

2

4

6

8

10

12

14

0 500 1000 1500 2000Vb

pH

Cb = 0.0001 MCb = 0.001 M

Cb = 0.01 M

pKa = 4.75Ca = 0.01 M


Page 60

Titration of a ___________ with a _________:

][][ +

+ ++

⋅=+

+ −+ HK

VVVC

VVVCH w

ba

aaA

ba

bbBH

αα

bBH

aA

aa

bb

COHH

COHH

VCVC

][][

][][

−+

−+

−+

−−

==Φ+

−

α

α

where +

+

+== +

++

BHBBH KH

HC

BH][

][][α ;

HA

HA

HAA KH

KCA

+== +

−

−

][][α

0

2

4

6

8

10

12

14


pH

Strong base Kb = 5

pKa = 4.75pKb = 5Ca = 0.01 M

0

2

4

6

8

10

12

14

0 5 10 15 20Vb

pH

Strong base

Kb = 5

pKa = 4.75pKb = 5Ca = 0.01 MVa = 10 mL


Page 61

Titration error

At the equivalence point ________

aabb VCVC ='

where Vb’ is Vb at the _______________

b

aab C

VCV ='

Titration error:

1111''

'.. −Φ=−=−=−=

−= ep

aa

epb

b

aa

ep

b

ep

b

bep

VCVC

CVC

VVV

VVV

eT

At the equivalence point and its vicinity:

b

ab

a

ba

CCC

VVV +

=+

Also, near the equivalence point, ____________ , thus _____________

aaa

a

KH

KHH

KHK ][

][][1

][1

+

+

+

+ −≅+

−=−+

=−α

−+

+−=−Φ ++ ][][

)(11 HHK

VCVV w

aa

baep α

a

w

aa

baep K

HHHK

VCVV ][][

][)(1

++

+ −

−+

=−Φ

a

epep

ep

w

ba

baep K

HH

HK

CCCC ][

][][

)(1+

++ −

−

+=−Φ


Page 62

Example: 0.1 M HAc, 0.1 M NaOH, pHep = __ instead of pH = _____ at theequivalence point

=−−=−Φ −

−−−

75.4

886

1010)1010)(20(1ep

Titration of a weak base with a strong acid:

ep

a

ep

wep

ba

baep H

KHKH

CCCC

][][][)(1 ++

+ −

−

+=−Φ

Análisis cualitativo

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